# What mass of oxygen gas is required for combustion of a 12.0*g mass of propane?

Apr 3, 2017

We need (i) a stoichiometric equation:

#### Explanation:

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$

Is it balanced? It's your problem not mine.

And (ii) we need equivalent quantities of propane and dioxygen.

$\text{Moles of propane} = \frac{12.0 \cdot g}{44.1 \cdot g \cdot m o {l}^{-} 1} = 0.272 \cdot m o l$

$\text{Moles of dioxygen} = \frac{5.01 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} = 0.156 \cdot m o l$

Clearly, there is insufficient oxygen for complete combustion: we would require $5 \times 0.272 \cdot m o l$ dioxygen gas.

So what would occur? Well, this is a case of incomplete combustion. I could invoke a stoichiometric equation of the sort:

${C}_{3} {H}_{8} \left(g\right) + 4 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + C + 4 {H}_{2} O \left(l\right)$

OR

${C}_{3} {H}_{8} \left(g\right) + \frac{7}{2} {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + C O \left(g\right) + C \left(s\right) + 4 {H}_{2} O \left(l\right)$

Which are both still deficient with respect to the oxidant. This question is NOT well-proposed; certainly it is unsuitable for an A-level student, and it would tax many first year university students.

You could assume COMPLETE combustion on the basis of the first equation, but the product mix would likely have $C$ and $C O$, the products of incomplete combustion. I take it you could work out the stoichiometric excess of propane on the basis of complete combustion? If not, raise a flag, and someone will readdress the problem.